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From: Deborah Ariel Pickett <debbiep@mail.csse.monash.edu.au>
Message-Id: <200201181054.g0IAs4S01786@bruce.csse.monash.edu.au>
Subject: Re: list of files with full path
To: guckes@math.fu-berlin.de (Sven Guckes)
Date: Fri, 18 Jan 2002 21:54:03 +1100 (EST)
Cc: zsh-users@sunsite.auc.dk (ZShell Users)
In-Reply-To: <20020118104317.GA23929@ritz.math.fu-berlin.de> from "Sven Guckes" at Jan 18, 2002 11:43:17 AM
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> I looked at the manual about "modifiers" -
> and they all seem to cut off something.
> now, what to use to make a listing of files
> with a full path?  is there a modifier for that?
> 
> $ ls **/*(???)
> /dir/file
> /dir/subdir1/file1
> /dir/subdir1/file2
> /dir/subdir2/file1
> /dir/subdir2/file2
> etc
> 
> There *must* be something for
> that with the ZShell, right?  ;-)

This works for me:

% print -l **/*

or, if you need to use ls, at least the -d option to show directories
rather than their contents:

% ls -d -1 **/*


If you want to prefix the current working directory, try this:

% print -l $PWD/**/*
or
% ls -d -1 $PWD/**/*

zsh doesn't strip paths from filenames unless you use one of the "word
modifiers" like ":t" ("tail"):

% print -l $PWD/**/*(:t)
(this won't be meaningful with ls)


If you want to print only files, and omit the directory names that
contain them (as your example above does, by leaving off /dir and
/dir/subdir{1,2}), the modifier is (.) for normal files only:

% ls -d -1 **/*(.)
or
% print -l **/*
or
% ls -d -1 $PWD/**/*(.)
or
% print -l $PWD/**/*(.)

I think the latter two produce the output you are seeking, based on your
example.


Or have I misunderstood the question?

-- 
Debbie Pickett http://www.csse.monash.edu.au/~debbiep debbiep@csse.monash.edu.au
"Where's the man, where's the child? Wrapped together side by side, who can tell
   you what to do when Mr Time has come for you?" - _Mr Time_, Alan Parsons